Q.1. An open organ pipe having fundamental frequency (n) is in unison with a vibrating string. If the tube is dipped in water so that 75% of the length of the tube is inside the water then the ratio of fundamental frequency of the air column of dipped tube with that of string will be (Neglect end corrections)
A. 1:1
B. 2:1
C. 2:3
D. 3:2
Answer :- B. 2:1
Explanation :-
\mathbf{n}_{\text {open }}=\frac{\mathrm{v}}{2 \mathrm{~L}} \ldots \ldots \text { (i) }When dipped in water, the pipe becomes closed at one end and open at the other.
Length available for resonance is
l_1=25 \% \times \mathrm{L} =\mathrm{L} \times \frac{25}{100} =\mathrm{L} / 4 \therefore \quad \mathrm{n}_{\text {closed }} \quad=\frac{\mathrm{v}}{4 l_1}=\frac{\mathrm{v}}{4 \times \frac{\mathrm{L}}{4}}=\frac{\mathrm{v}}{\mathrm{~L}}Comparing (i) and (ii),
\therefore \quad \frac{\mathrm{n}_{\text {closed }}}{\mathrm{n}_{\text {open }}}=\frac{\left(\frac{\mathrm{v}}{\mathrm{~L}}\right)}{\left(\frac{\mathrm{v}}{2 \mathrm{~L}}\right)}=\frac{2}{1}The ratio of fundamental frequency of the air column of dipped tube with that of string will be 2:1. Therefore, option (B) is the correct answer.